[Guest]

my friend was asking me this, but i cannot find a simple solution to describe it, and im confident i have the right answer i just cant explain it to him

there is 26 letters in the alphabet

how many 3 letter variations are there

eg

AAA

or AAB

or APQ

or XTF

etc..

can anyone help me explain it simply

btw i know/think its

26*26*26

[Guest]

There are 26 letters in the alphabet, so obviously there are 26 possible combinations of one letter. To each of these 26 one-letter combinations you could add any of 26 letters to make it a two-letter combination. So, for each of the original 26 letters, there are 26 combinations. Multiply 26 by 26 to get the number of two-letter combinations.

The same logic applies for the third letter. For each of the 26*26 possible two-letter combinations, you could add any of 26 letters to make it a three-letter combination. So, once again, multiply by 26.

Your answer, 26*26*26, is correct.

[Guest]

I learned this in math haha =D

Like Seventh Sage has said, each letter has 26 possible letters. There are 26 possible letters in 3 different step. So along with the Counting Principle, all you do is multiply the number of options for each step.

Hence, 26*26*26, or 26 cubed, if you want to be fast.

[Guest]

I just thought of something. Does order matter? That is, do the combination ABC, CBA, BCA, ACB, BAC, CAB count as different combinations, or are all those the same in this problem. If they count as the same combination of three letters, that changes everything.

[Guest]

small case, = 26*2*26*2*26*2

Edited June 15, 2008 by magicalcheese

[Guest]

it is given a set A of n elements.

1. Subsets with j elements (j<=n) of A are called Combinations of n taken by j. C= n!/j!(n-j)! where n!=1*2*3*3*...*(n-1)*n.

2.Subsets in which order matters, subsets with j elements of A are called Arrangements of n taken by j. A= n!/(n-j)!. In this case, n=26, j=3;

Edited June 15, 2008 by pinkfloyd

[Guest]

I just thought of something. Does order matter? That is, do the combination ABC, CBA, BCA, ACB, BAC, CAB count as different combinations, or are all those the same in this problem. If they count as the same combination of three letters, that changes everything.

This is complex, because we can have none, one or two repeating letters.

For reference, a permutation is the arrangements possible.

For case 1, no repeats, we use nPr on the calculator, which finds permutations. And we get from 26nPr3 = 15600. (remember that this does not calculate repeats) We can divide by six (since there are six permutations of the same combination)

Or to even simply solve case one, we can use 26nCr3 which finds 2600 automatically.

For case 2, with one repeat, the first letter can be anything, and then the second letter must be the same, the third can be any other of the 25, so we get: 26*1*25 = 650

(Each of these in case two have three permutations: AAB, ABA, BAA, but we do not need to divide by three this time because 26*1*25 should only calculate one of the possible permutations.)

For case 3, all repeats there are 26 of them: AAA, BBB, CCC... etc.

So when we add up all these cases we get the total number of combinations.

3276?

Can someone double check my math?

[Guest]

I just thought of something. Does order matter? That is, do the combination ABC, CBA, BCA, ACB, BAC, CAB count as different combinations, or are all those the same in this problem. If they count as the same combination of three letters, that changes everything.

Are you meaning ????

A A A

A A A

A A A

A A A

A A A

A A A

[Guest]

Are you meaning ????

A A A

A A A

A A A

A A A

A A A

A A A

you can only have one set of AAA's, treat it like a number plate.

eg how many number plates can there be in london under the same year.

[Guest]

Are you meaning ????

A A A

A A A

A A A

A A A

A A A

A A A

No those are all the same thing.

He means: how many different possibilities could you get?

EX.

AAA

AAB

AAC

AAD

ect.

BAA

BAB

BAC

BAD

ect.

CAA

CAB

CAC

CAD

ect.

and so on... until you get to the z's

there is only one 'AAA'.

[Guest]

No those are all the same thing.

He means: how many different possibilities could you get?

EX.

AAA

AAB

AAC

AAD

ect.

BAA

BAB

BAC

BAD

ect.

CAA

CAB

CAC

CAD

ect.

and so on... until you get to the z's

there is only one 'AAA'.

Can you actually answer this on behalf of the OP starter?

You can give your answer and reasoning/solution

Besides a WORD that has a letter order changed is not the same WROD

just as in crib with priles and pairs when holding four of a kind three priles and six pairs (suit irrelevant here - it's a number count)

[Guest]

One way to look at is to consider that you are working with a base 26 number system, using the letters of the alphabet as digits (a=0, b=1,...,z=25). As there are 1000 numbers from 0~999 in decimal, there must be BAAA numbers from AAA~ZZZ in base 26. So your question is the same as "Convert the base 26 number BAAA to decimal". And the answer is 17576. (Because BAAA = 1x26^3 + 0x26^2 + 0x26^1 + 0x26^0 = 26^3 = 17576.)

[Yoruichi-san]

I like oranfry's method a lot. But in case any one cares, here is how to do it the conventional way with combinatorics:

Okay, so as said before, there are 3 cases to consider:

Case I: No repeating letters. (i.e. ABC)

Case II: 1 repeating letter (i.e. AAB)

Case III: 2 repeating letters (i.e. AAA)

The total # of combinations is the sum of these three cases. Here is how I look at them:

Case I: No repeating letters means that all three letters are different. Therefore, there are 26 choices for the first letter, 25 choices for the second, and 24 choices for the third. So there are a total of 26*25*24=15600 three-letter words of case I. Note that order *does* matter here, i.e. ABC is different from ACB. This is a permutation, not a combination. Put in mathematical terms, this is 26P3 (26 permutation 3).

Case II: This is the tricky one. The way I think about it is that each of these has two unique letters, and the third letter is a repeat of one of the two.

So first I figure out the number of two-letter combinations (i.e. AB), which is 26C2=26*25/2=325.

Now for each of these two-letter combinations, there are 2 choices for the third letter (i.e. we can have a repeating A or B) and 3 places to put the non-repeating letter, so we can make 2*3=6 different three-letter words, (i.e. AAB, ABA, BAA, BBA, BAB, ABB).

So there are a total of 6*325=1950 total words for case II.

Case III: This is an easy one. Only one letter to choose, so there are 26 (or 26C1) choices for the letter. 26 words for case III.

So there are a total of 15600+1950+26=17576 possible words of length three.

[Guest]

I like oranfry's method a lot. But in case any one cares, here is how to do it the conventional way with combinatorics:

Okay, so as said before, there are 3 cases to consider:

Case I: No repeating letters. (i.e. ABC)

Case II: 1 repeating letter (i.e. AAB)

Case III: 2 repeating letters (i.e. AAA)

The total # of combinations is the sum of these three cases. Here is how I look at them:

Case I: No repeating letters means that all three letters are different. Therefore, there are 26 choices for the first letter, 25 choices for the second, and 24 choices for the third. So there are a total of 26*25*24=15600 three-letter words of case I. Note that order *does* matter here, i.e. ABC is different from ACB. This is a permutation, not a combination. Put in mathematical terms, this is 26P3 (26 permutation 3).

Case II: This is the tricky one. The way I think about it is that each of these has two unique letters, and the third letter is a repeat of one of the two.

So first I figure out the number of two-letter combinations (i.e. AB), which is 26C2=26*25/2=325.

Now for each of these two-letter combinations, there are 2 choices for the third letter (i.e. we can have a repeating A or B) and 3 places to put the non-repeating letter, so we can make 2*3=6 different three-letter words, (i.e. AAB, ABA, BAA, BBA, BAB, ABB).

So there are a total of 6*325=1950 total words for case II.

Case III: This is an easy one. Only one letter to choose, so there are 26 (or 26C1) choices for the letter. 26 words for case III.

So there are a total of 15600+1950+26=17576 possible words of length three.

There's a significantly simpler conventional method, where order doesn't matter.

This method is often referred to as "stars and bars".

Think of having 26 "drawers" (labeled A through Z) and 3 undistinguished "objects" that can be placed in them. We interpret, for example, having 2 objects in the drawer labeled "X" as having two occurrences of the letter X.

How do we figure out how many ways there are to place objects in drawers? We represent an object by a * (star), and the division between drawers by a | (bar), and determine how many possible strings there are of 3 *s and 25 |s. (Why 25? There are 26 drawers, so 25 divisions between them.)

With 28 total symbols, there are 28! (28 factorial, 28*27*26*...*2*1) possible strings. But many of these are repeats: 3! ways of arranging the stars, 25! ways of arranging the bars. So the number of sets of three letters is

28!/(3!*25!)=28C3=3276

This is as opposed to 26^3=17576 if order matters.

If order does matter, the combinatorial method is the easy one posted as the first solution.

Edited June 17, 2008 by lytefoot

[Guest]

Lost in Space i confirm that that is what i meant.

we first had the arguement over number plates and 2 number plates cant have AAA even though 2 or 3 of the A's are switched around

sorry i should have put that in at the start

[Guest]

I just thought of something. Does order matter? That is, do the combination ABC, CBA, BCA, ACB, BAC, CAB count as different combinations, or are all those the same in this problem. If they count as the same combination of three letters, that changes everything.

Ok some combinatoriks.

They are called varations of n elements of k clas - those are the varations of A, B and C. But AAA is not a varation, so all varations are 26*25*24 or n!/(n-k)! . You may look for combinations only witch is (n!/(n-k))/k! (varations/permutations - those are all the varations when k=0 and are = to n!) - . So they are 22100 combinations and 132 600 varations. Hope happy